+26393 
$$\small{\text{$
\begin{array}{rcl}
\boxed{~~ax^2+bx+c=0 \qquad
x_{1,2} = \dfrac{ -b\pm \sqrt{b^2-4ac} }{2a} ~~}
\end{array}
$}}$$
$$\small{\text{$
\begin{array}{rcl}
a=m ~~b=-1~~c=-\dfrac{m-2}{4}\\\\
x_{1,2} &=& \dfrac{ -b\pm \sqrt{b^2-4ac} }{2a} \\\\
x_{1,2} &=& \dfrac{ 1\pm \sqrt{1^2-4m\left(-\dfrac{m-2}{4}\right) } }{2m} \\\\
x_{1,2} &=& \dfrac{ 1\pm \sqrt{1+m\left(m-2\right) } }{2m} \\\\
x_{1,2} &=& \dfrac{ 1\pm \sqrt{1+m^2-2m} }{2m} \\\\
x_{1,2} &=& \dfrac{ 1\pm \sqrt{1-2m+m^2} }{2m} \\\\
x_{1,2} &=& \dfrac{ 1\pm \sqrt{(1-m)^2} }{2m} \\\\
x_{1,2} &=& \dfrac{ 1\pm (1-m) }{2m} \\\\
x_1 &=& \dfrac{ 1 + (1-m) }{2m} \\\\
x_1 &=& \dfrac{ 2-m }{2m} \\\\
x_1 &=& \dfrac{ 2 }{2m} - \dfrac{ m }{2m} \\\\
\mathbf{x_1} & \mathbf{=} & \mathbf{ \dfrac{ 1 }{m} - \dfrac{ 1 }{2} } \\\\\\
x_2 &=& \dfrac{ 1 - (1-m) }{2m} \\\\
x_2 &=& \dfrac{ 1 - 1 + m)}{2m} \\\\
x_2 &=& \dfrac{ m }{2m} \\\\
\mathbf{x_2} & \mathbf{=} & \mathbf{\dfrac{ 1 }{2}}
\end{array}
$}}$$
+14538 $${{\mathtt{mx}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\frac{\left({\mathtt{m}}{\mathtt{\,-\,}}{\mathtt{2}}\right)}{{\mathtt{4}}}} = {\mathtt{0}}$$
+33661 geno has already answered this here: http://web2.0calc.com/questions/help-me-please_38
.
+118673 Sabi92, if you do not understand Geno's answer, you need to post questions about it on the original thread. :)
+26393
$$\small{\text{$
\begin{array}{rcl}
\boxed{~~ax^2+bx+c=0 \qquad
x_{1,2} = \dfrac{ -b\pm \sqrt{b^2-4ac} }{2a} ~~}
\end{array}
$}}$$
$$\small{\text{$
\begin{array}{rcl}
a=m ~~b=-1~~c=-\dfrac{m-2}{4}\\\\
x_{1,2} &=& \dfrac{ -b\pm \sqrt{b^2-4ac} }{2a} \\\\
x_{1,2} &=& \dfrac{ 1\pm \sqrt{1^2-4m\left(-\dfrac{m-2}{4}\right) } }{2m} \\\\
x_{1,2} &=& \dfrac{ 1\pm \sqrt{1+m\left(m-2\right) } }{2m} \\\\
x_{1,2} &=& \dfrac{ 1\pm \sqrt{1+m^2-2m} }{2m} \\\\
x_{1,2} &=& \dfrac{ 1\pm \sqrt{1-2m+m^2} }{2m} \\\\
x_{1,2} &=& \dfrac{ 1\pm \sqrt{(1-m)^2} }{2m} \\\\
x_{1,2} &=& \dfrac{ 1\pm (1-m) }{2m} \\\\
x_1 &=& \dfrac{ 1 + (1-m) }{2m} \\\\
x_1 &=& \dfrac{ 2-m }{2m} \\\\
x_1 &=& \dfrac{ 2 }{2m} - \dfrac{ m }{2m} \\\\
\mathbf{x_1} & \mathbf{=} & \mathbf{ \dfrac{ 1 }{m} - \dfrac{ 1 }{2} } \\\\\\
x_2 &=& \dfrac{ 1 - (1-m) }{2m} \\\\
x_2 &=& \dfrac{ 1 - 1 + m)}{2m} \\\\
x_2 &=& \dfrac{ m }{2m} \\\\
\mathbf{x_2} & \mathbf{=} & \mathbf{\dfrac{ 1 }{2}}
\end{array}
$}}$$
+14538
!
+118673 Why didn't you see his answer?
The fact that you had an answer would have been clearly displayed in your watch list. :/
+262 yes but i think i confused it with other answer and reloaded the browser whithout noticed that its exactly the answer i need
